Problem: The volume of a cylinder is increasing at a rate of $10\pi$ cubic meters per hour. The height of the cylinder is fixed at $5$ meters. At a certain instant, the volume is $80\pi$ cubic meters. What is the rate of change of the surface area of the cylinder at that instant (in square meters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{4}$ (Choice B) B $\dfrac{26\pi}{3}$ (Choice C) C $\dfrac{13\pi}{2}$ (Choice D) D $\sqrt[3]{100\pi^2}$ The surface area of a cylinder with base radius $r$ and height $h$ is $2\pi r^2+2\pi r h$. The volume of a cylinder with base radius $r$ and height $h$ is $\pi r^2h$.
Setting up the math Let... $r(t)$ denote the cylinder's base radius at time $t$, $h$ denote the cylinder's height (which is always $5$ meters), $V(t)$ denote the cylinder's volume at time $t$, and $S(t)$ denote the cylinder's surface area at time $t$. We are given that $h=5$ and $V'(t)=10\pi$, We are also given that $V(t_0)=80\pi$ for a specific time $t_0$. We want to find $S'(t_0)$. Relating the measures $S(t)$ and $r(t)$ relate to each other through the formula for the surface area of a cylinder: $\begin{aligned} S(t)&=2\pi [r(t)]^2+2\pi r(t)h \\\\ &=2\pi[r(t)]^2+10\pi r(t) \end{aligned}$ We can differentiate both sides to find an expression for $S'(t)$ : $S'(t)=4\pi r(t)r'(t)+10\pi r'(t)$ $V(t)$ and $r(t)$ relate to each other through the formula for the volume of a cylinder: $\begin{aligned} V(t)&=\pi[r(t)]^2h \\\\ &=5\pi[r(t)]^2 \end{aligned}$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=10\pi r(t)r'(t)$ Using the information to solve Let's plug ${V(t_0)}={80\pi}$ into the expression for $V(t_0)$ : $\begin{aligned} {V(t_0)}&=5\pi[r(t_0)]^2 \\\\ {80\pi}&=5\pi[r(t_0)]^2 \\\\ 16&=[r(t_0)]^2 \\\\ {4}&={r(t_0)} \end{aligned}$ Let's plug ${V'(t_0)}={10\pi}$ and ${r(t_0)}={4}$ into the expression for $V'(t_0)$ : $\begin{aligned} {V'(t_0)}&=10\pi{r(t_0)}r'(t) \\\\ {10\pi}&=10\pi( 4)r'(t_0) \\\\ C{\dfrac14}&=C{r'(t_0)} \end{aligned}$ Now let's plug ${r(t_0)}={4}$ and $C{r'(t_0)}=C{\dfrac{1}{4}}$ into the expression for $S'(t_0)$ : $\begin{aligned} S'(t_0)&=4\pi{r(t_0)}C{r'(t_0)}+10\piC{r'(t_0)} \\\\ &=4\pi({4})\left(C{\dfrac{1}{4}}\right)+10\pi\left(C{\dfrac{1}{4}}\right) \\\\ &=\dfrac{13\pi}{2} \end{aligned}$ In conclusion, the rate of change of the surface area of the cylinder at that instant is $\dfrac{13\pi}{2}$ square meters per hour. Since the rate of change is positive, we know that the surface area is increasing.